A `5.0 mL` of solution of `H_(2)O_(2)` liberates `0.508 g` of iodine from acidified `KI` solution. Calculate the strength of `H_(2)O_(2)` solution in terms of volume strength at `STP`.

Explanation

m" Eq of "`I_(2)=(0.508)/(127)xx1000`
m" Eq of "`KI=(0.508)/(127)xx1000`
m" Eq of "`H_(2)O_(2)=(0.508)/(127)xx1000`
Normality of `H_(2)O_(2)=(0.508)/(127)xx(1000)/(5)`
for `H_(2)O_(2)`,
`5.6xx` Normality`=` Volume strength
`therefore` Volume strength`=(0.508)/(127)xx(1000)/(5)xx5.6=4.48V`